Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9 + 7;
const int N = 3e5 + 10;
vector<pair<double, double>> vp;
int sum[N];
double check(int x1, int y1, int x2, int y2) {
  if (x1 == x2)
    return x1;
  else
    return -1.0 * y1 * (x2 - x1) / (y2 - y1) + x1;
}
double combine(double l1, double r1, double l2, double r2) {
  return max(0.0, min(r2, r1) - max(l1, l2));
}
double getans(double x) {
  int o = lower_bound(vp.begin(), vp.end(), make_pair(x, -1.0)) - vp.begin();
  return sum[o] + combine(vp[o].second, vp[o].first, 0, x);
}
int main() {
  int sy, a, b;
  scanf("%d%d%d", &sy, &a, &b);
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    sum[i + 1] = sum[i] + (r - l);
    vp.push_back({1.0 * r, 1.0 * l});
  }
  vp.push_back({1.0 * INF * INF, 1.0 * INF * INF});
  int m;
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    double x1 = check(x, y, a, sy);
    double x2 = check(x, y, b, sy);
    double len = getans(x2) - getans(x1);
    printf("%.15f\n", len * (y - sy) / y);
  }
  return 0;
}